44). In a trapezium, the two non-parallel sides are equal in length, each being of 5 units. The parallel sides are at a distance of 3 units apart. If the smaller side of the parallel sides is of length 2 units, then the sum of the diagonals of the trapezium is

A). \( \Large 10\sqrt{5} \) units

B). \( \Large 6\sqrt{5} \) units

C). \( \Large 5\sqrt{5} \) units

D). \( \Large 3\sqrt{5} \) units

View Answer

Correct Answer: \( \Large 6\sqrt{5} \) units

In \( \Large \triangle BCF \), by Pythagoras theorem,

\( \Large (5)^{2}=(3)^{2}+(BF)^{2} \)

BF=4 cm

AB=2+4+4=10 cm

Now, in \( \Large \triangle ACF \), \( \Large AC^{2}=CF^{2}+FA^{2} \)

=> \( \Large AC^{2}=3^{2}+6^{2} \) => AC=\( \Large \sqrt{45} \)cm

Similarly, BD = \( \Large \sqrt{45} \)cm

Sum of diagonals

=AC+BD= \( \Large \sqrt{45}+\sqrt{45}=2\sqrt{45}=6\sqrt{5} \)cm

45). The ratio of the length of the parallel sides of a trapezium is 3 : 2. The shortest distance between them is 15 cm. If the area of the trapezium is 450 sq cm, find the sum of the lengths of the parallel sides.